Rebalancing Frequency Analysis

Continuous delta hedging is an idealization that cannot be achieved in practice. Real hedging occurs at discrete times, introducing a random hedging error. This section analyzes how the rebalancing frequency affects hedging quality, derives the fundamental scaling laws, and studies the cost-error tradeoff that arises when transaction costs are included.

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Setup: Discrete Hedging Framework

The Hedging Problem

Consider an option \(V(t,S)\) hedged by holding \(\Delta_k = \Delta(t_k, S_{t_k})\) shares of the underlying at discrete rebalancing times \(t_0 < t_1 < \cdots < t_N = T\) with uniform spacing \(\delta t = T/N\).

The hedge portfolio evolves as:

\[ \Pi_{k+1} = \Pi_k\, e^{r\,\delta t} + \Delta_k\left(S_{k+1} - S_k\, e^{r\,\delta t}\right) \]

The hedging error at each step is the mismatch between the option value change and the hedge portfolio change:

\[ \epsilon_k = \left(V_{k+1} - V_k\right) - \left(\Pi_{k+1} - \Pi_k\right) \]

Taylor Expansion of the Error

Using the Ito-Taylor expansion (ignoring discounting for clarity):

\[ V_{k+1} - V_k \approx \Delta_k\,\delta S_k + \Theta_k\,\delta t + \frac{1}{2}\Gamma_k\,(\delta S_k)^2 \]

where \(\delta S_k = S_{k+1} - S_k\). Since the hedge captures only \(\Delta_k\,\delta S_k\):

\[ \boxed{\epsilon_k \approx \Theta_k\,\delta t + \frac{1}{2}\Gamma_k\,(\delta S_k)^2} \]

Using the theta-gamma identity \(\Theta + \frac{1}{2}\sigma^2 S^2 \Gamma \approx r(V - S\Delta)\):

\[ \epsilon_k \approx \frac{1}{2}\Gamma_k\left[(\delta S_k)^2 - \sigma^2 S_k^2\,\delta t\right] \]

This reveals that the hedging error is driven by the difference between realized and expected squared price moves.

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Variance of the Hedging Error

Single-Step Variance

Under Black-Scholes dynamics with \(\delta S_k = \sigma S_k \sqrt{\delta t}\, Z_k + O(\delta t)\) where \(Z_k \sim \mathcal{N}(0,1)\):

\[ (\delta S_k)^2 = \sigma^2 S_k^2\,\delta t\, Z_k^2 + O(\delta t^{3/2}) \]

The conditional mean and variance of \(\epsilon_k\) are:

\[ \mathbb{E}[\epsilon_k \mid \mathcal{F}_{t_k}] \approx 0 \]

\[ \operatorname{Var}(\epsilon_k \mid \mathcal{F}_{t_k}) \approx \frac{1}{2}\Gamma_k^2 S_k^4 \sigma^4\,(\delta t)^2 \]

The factor \(\frac{1}{2}\) comes from \(\operatorname{Var}(Z^2) = 2\) for \(Z \sim \mathcal{N}(0,1)\).

Cumulative Error Variance

Summing over \(N = T/\delta t\) independent steps:

\[ \boxed{\operatorname{Var}(\mathrm{HE}) \approx \frac{1}{2}\,\overline{\Gamma^2 S^4 \sigma^4}\, T\,\delta t} \]

where \(\overline{\Gamma^2 S^4 \sigma^4}\) denotes the time-averaged quantity. The key scaling law is:

\[ \operatorname{Std}(\mathrm{HE}) \propto \sqrt{\delta t} = \frac{1}{\sqrt{N}} \cdot \sqrt{T} \]

Theorem (Hedging Error Scaling). Under Black-Scholes dynamics with \(N\) equally spaced rebalancing dates, the standard deviation of the cumulative hedging error satisfies:

\[ \operatorname{Std}(\mathrm{HE}) = c_\Gamma \cdot \sqrt{\frac{T}{N}} \]

where \(c_\Gamma = \sqrt{\frac{1}{2}\int_0^T \mathbb{E}[\Gamma(t,S_t)^2 S_t^4 \sigma^4]\,dt}\) is a constant depending on the option characteristics.

Interpretation

• Doubling the rebalancing frequency reduces hedging error standard deviation by a factor of \(\sqrt{2} \approx 1.41\).
• To halve the hedging error, you need four times as many rebalancing dates.
• This \(\sqrt{N}\) convergence is a consequence of the central limit theorem applied to the sum of per-step errors.

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Asymptotic Distribution

Central Limit Theorem for Hedging Error

For large \(N\), the cumulative hedging error is approximately Gaussian:

\[ \mathrm{HE}_N \xrightarrow{d} \mathcal{N}\!\left(0,\; \frac{1}{2}\int_0^T \Gamma^2 S_t^4 \sigma^4\,dt \cdot \delta t\right) \]

More precisely, the normalized hedging error converges:

\[ \frac{\mathrm{HE}_N}{\sqrt{\delta t}} \xrightarrow{d} \mathcal{N}\!\left(0,\; \frac{1}{2}\int_0^T \Gamma^2 S_t^4 \sigma^4\,dt\right) \]

This result, due to Bertsimas, Kogan, and Lo (2000), provides the basis for confidence intervals:

\[ \Pr\!\left(|\mathrm{HE}_N| \leq 1.96 \cdot \operatorname{Std}(\mathrm{HE})\right) \approx 0.95 \]

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Frequency-Error Table

Consider an ATM call with \(S = K = 100\), \(\sigma = 20\%\), \(T = 0.25\) years, \(r = 5\%\). The gamma is approximately \(\Gamma \approx 0.032\), giving \(c_\Gamma \approx \frac{1}{\sqrt{2}} \times 0.032 \times 10{,}000 \times 0.04 \times 0.5 \approx 4.5\).

Frequency	\(N\)	\(\delta t\) (years)	\(\operatorname{Std}(\mathrm{HE})\)	Relative to daily
Monthly	3	1/12	\(\sim \$1.30\)	3.2x
Weekly	13	1/52	\(\sim \$0.63\)	1.5x
Daily	63	1/252	\(\sim \$0.40\)	1.0x
Twice daily	126	1/504	\(\sim \$0.28\)	0.71x
Hourly	504	1/2016	\(\sim \$0.14\)	0.35x
Continuous	\(\infty\)	0	0	0

The pattern is clear: moving from weekly to daily reduces the error by about \(\sqrt{5} \approx 2.2\) times; moving from daily to hourly reduces it by about \(\sqrt{8} \approx 2.8\) times.

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Transaction Costs and the Cost-Error Tradeoff

Transaction Cost Model

Each rebalancing incurs a cost proportional to the number of shares traded:

\[ \text{TC}_k = \kappa\, S_k\, |\Delta_{k+1} - \Delta_k| \]

where \(\kappa\) is the proportional transaction cost (half-spread). The total transaction cost over \([0, T]\) is:

\[ \text{TC}_{\text{total}} = \kappa \sum_{k=0}^{N-1} S_k\, |\Delta_{k+1} - \Delta_k| \]

Expected Transaction Cost

The delta change at each step satisfies:

\[ |\Delta_{k+1} - \Delta_k| \approx |\Gamma_k|\,|\delta S_k| \approx |\Gamma_k|\, \sigma S_k\, \sqrt{\delta t}\, |Z_k| \]

Taking expectations (using \(\mathbb{E}[|Z|] = \sqrt{2/\pi}\)):

\[ \mathbb{E}[\text{TC}_k] \approx \kappa\, |\Gamma_k|\, \sigma S_k^2 \sqrt{\frac{2\,\delta t}{\pi}} \]

Summing over \(N\) steps:

\[ \boxed{\mathbb{E}[\text{TC}_{\text{total}}] \approx \kappa\, \overline{|\Gamma| \sigma S^2}\, T \sqrt{\frac{2}{\pi\,\delta t}} = C_0 \sqrt{N}} \]

where \(C_0\) depends on the option parameters and spread. Transaction costs grow as \(\sqrt{N}\) --- more frequent rebalancing is increasingly expensive.

The Tradeoff

Component	Scaling with \(N\)
Hedging error std	\(\propto 1/\sqrt{N}\) (decreasing)
Expected transaction cost	\(\propto \sqrt{N}\) (increasing)
Total cost (error + TC)	\(\propto 1/\sqrt{N} + c\sqrt{N}\) (U-shaped)

The total cost function is:

\[ \text{Total}(N) = \frac{a}{\sqrt{N}} + b\sqrt{N} \]

where \(a\) represents the hedging error impact and \(b\) represents the transaction cost per rebalancing.

Optimal Rebalancing Frequency

Minimizing the total cost:

\[ \frac{d}{dN}\text{Total}(N) = -\frac{a}{2N^{3/2}} + \frac{b}{2\sqrt{N}} = 0 \]

\[ \boxed{N^* = \frac{a}{b} = \frac{\text{error sensitivity}}{\text{transaction cost rate}}} \]

The optimal frequency is:

\[ \delta t^* = T \cdot \frac{b}{a} \]

Practical Interpretation

• Liquid markets (small \(\kappa\), small \(b\)): Rebalance frequently (\(N^*\) is large).
• Illiquid markets (large \(\kappa\), large \(b\)): Rebalance less often.
• High gamma (large \(a\)): More frequent rebalancing is warranted.
• Low gamma (small \(a\)): The hedging error is small regardless, so less frequent rebalancing suffices.

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Leland's Adjusted Volatility

Modified Hedging Strategy

Leland (1985) proposed adjusting the implied volatility used for computing deltas to account for transaction costs. The Leland volatility is:

\[ \hat{\sigma}^2 = \sigma^2\left(1 + \sqrt{\frac{2}{\pi}} \cdot \frac{\kappa}{\sigma\sqrt{\delta t}}\right) \]

By hedging with \(\hat{\Delta} = \Delta(\hat{\sigma})\) instead of \(\Delta(\sigma)\), the expected hedging cost (error plus transaction costs) is reduced. The adjustment widens the effective volatility, which systematically over-hedges to compensate for the discrete rebalancing.

Interpretation

The Leland adjustment adds a correction term of order \(O(1/\sqrt{\delta t})\) to the volatility. In the limit \(\delta t \to 0\) (continuous hedging), \(\hat{\sigma} \to \sigma\) and the adjustment vanishes. For finite \(\delta t\), the adjustment accounts for the asymmetric impact of transaction costs on the hedging P&L.

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Gamma-Weighted Rebalancing

Non-Uniform Schedules

The analysis above assumes uniform rebalancing. Since gamma varies over the option's life (peaking near expiry for ATM options), a non-uniform schedule can improve efficiency.

Principle: Rebalance more frequently when gamma is large, less frequently when gamma is small.

A simple adaptive rule: rebalance when the estimated hedging error exceeds a threshold:

\[ \frac{1}{2}|\Gamma_k| \cdot \sigma^2 S_k^2 \cdot (t - t_k) > \varepsilon_{\text{target}} \]

This leads to shorter intervals near the strike and near expiry, and longer intervals when the option is deep in- or out-of-the-money.

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Numerical Example

Consider delta-hedging a short ATM European call (\(S = K = 100\), \(\sigma = 25\%\), \(T = 0.5\), \(r = 3\%\)) with transaction costs \(\kappa = 0.001\) (10 bps half-spread).

Parameters:

• Option price \(\approx \$7.50\)
• Average \(\Gamma S^2 \sigma^2 / 2 \approx 20\) (dollar gamma)
• Transaction cost per rebalance \(\approx \$0.05\)

Comparison of frequencies:

Frequency	\(N\)	HE Std	Expected TC	Total
Weekly	26	$0.98	$0.26	$1.24
Daily	126	$0.45	$0.56	$1.01
Twice daily	252	$0.32	$0.79	$1.11
Hourly	1008	$0.16	$1.59	$1.75

The minimum total cost occurs near daily rebalancing for this set of parameters. Hourly rebalancing improves hedging accuracy but the transaction costs dominate. Weekly rebalancing saves on costs but leaves too much hedging error.

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Higher-Order Corrections

Beyond the Leading Term

The \(\sqrt{\delta t}\) scaling captures the leading-order behavior. Higher-order asymptotic expansions reveal:

\[ \operatorname{Var}(\mathrm{HE}) = c_1\,\delta t + c_2\,(\delta t)^2 + O((\delta t)^3) \]

where \(c_1\) is the gamma-squared integral and \(c_2\) involves the speed (derivative of gamma with respect to spot) and the charm (derivative of delta with respect to time). These corrections become relevant for very coarse rebalancing schedules.

Skewness

The hedging error distribution is not perfectly Gaussian. The leading-order skewness is:

\[ \operatorname{Skew}(\mathrm{HE}) \propto (\delta t)^{1/2} \]

which vanishes as \(\delta t \to 0\), consistent with the CLT. For finite \(\delta t\), the distribution is slightly right-skewed for long gamma positions (since \((Z^2 - 1)\) has positive skewness).

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Summary

Result	Formula
Per-step error	\(\epsilon_k \approx \frac{1}{2}\Gamma_k[(\delta S_k)^2 - \sigma^2 S_k^2\,\delta t]\)
Error std scaling	\(\operatorname{Std}(\mathrm{HE}) \propto \sqrt{\delta t} = O(1/\sqrt{N})\)
Transaction cost scaling	\(\mathbb{E}[\text{TC}] \propto \sqrt{N}\)
Optimal frequency	\(N^* = a/b\) (error sensitivity / cost rate)
Asymptotic distribution	Gaussian by CLT
Leland adjustment	\(\hat{\sigma}^2 = \sigma^2(1 + \kappa\sqrt{2/(\pi\sigma^2\delta t)})\)
Key insight	Hedging quality is limited by the \(\sqrt{\delta t}\) barrier; transaction costs create a U-shaped total cost curve

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Exercises

Exercise 1. For an ATM call with \(S = K = 100\), \(\sigma = 0.25\), \(T = 0.5\), and constant \(\Gamma = 0.028\), compute the cumulative hedging error standard deviation for \(N = 26\) (weekly), \(N = 126\) (daily), and \(N = 504\) (twice daily) rebalancing dates. Verify that the ratios of these standard deviations are consistent with the \(\sqrt{\delta t}\) scaling law.

Solution to Exercise 1

With \(S = K = 100\), \(\sigma = 0.25\), \(T = 0.5\), \(\Gamma = 0.028\):

The cumulative hedging error variance is:

\[ \operatorname{Var}(\text{HE}) = \frac{1}{2}\Gamma^2 S^4 \sigma^4 T \cdot \delta t \]

First compute the constant factor:

\[ \frac{1}{2}\Gamma^2 S^4 \sigma^4 T = \frac{1}{2}(0.028)^2(10^8)(0.25)^4(0.5) \]

\[ = \frac{1}{2}(7.84 \times 10^{-4})(10^8)(3.906 \times 10^{-3})(0.5) \]

\[ = \frac{1}{2}(7.84 \times 10^{-4})(10^8)(1.953 \times 10^{-3}) = \frac{1}{2}(7.84 \times 10^{-4})(1.953 \times 10^{5}) \]

\[ = \frac{1}{2}(153.1) = 76.55 \]

Weekly (\(N = 26\), \(\delta t = T/N = 0.5/26 \approx 0.01923\)):

\[ \operatorname{Std}(\text{HE}) = \sqrt{76.55 \times 0.01923} = \sqrt{1.472} \approx 1.213 \]

Daily (\(N = 126\), \(\delta t = 0.5/126 \approx 0.003968\)):

\[ \operatorname{Std}(\text{HE}) = \sqrt{76.55 \times 0.003968} = \sqrt{0.3038} \approx 0.551 \]

Twice daily (\(N = 504\), \(\delta t = 0.5/504 \approx 0.000992\)):

\[ \operatorname{Std}(\text{HE}) = \sqrt{76.55 \times 0.000992} = \sqrt{0.07594} \approx 0.276 \]

Verification of \(\sqrt{\delta t}\) scaling:

\[ \frac{\text{Std(weekly)}}{\text{Std(daily)}} = \frac{1.213}{0.551} \approx 2.20 \approx \sqrt{126/26} = \sqrt{4.846} \approx 2.20 \;\checkmark \]

\[ \frac{\text{Std(daily)}}{\text{Std(twice daily)}} = \frac{0.551}{0.276} \approx 2.00 \approx \sqrt{504/126} = \sqrt{4} = 2.00 \;\checkmark \]

The ratios confirm the \(\sqrt{\delta t}\) scaling law.

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Exercise 2. The total cost function is \(\text{Total}(N) = a/\sqrt{N} + b\sqrt{N}\). Derive the optimal \(N^*\) that minimizes this function. For \(a = 4.0\) and \(b = 0.05\), compute \(N^*\) and the corresponding optimal rebalancing interval \(\delta t^*\) when \(T = 0.25\). What is the minimum total cost?

Solution to Exercise 2

The total cost function is:

\[ \text{Total}(N) = \frac{a}{\sqrt{N}} + b\sqrt{N} \]

Taking the derivative and setting it to zero:

\[ \frac{d}{dN}\text{Total}(N) = -\frac{a}{2N^{3/2}} + \frac{b}{2\sqrt{N}} = 0 \]

Multiplying through by \(2N^{3/2}\):

\[ -a + bN = 0 \implies N^* = \frac{a}{b} \]

For \(a = 4.0\) and \(b = 0.05\):

\[ N^* = \frac{4.0}{0.05} = 80 \]

The optimal rebalancing interval with \(T = 0.25\):

\[ \delta t^* = \frac{T}{N^*} = \frac{0.25}{80} = 0.003125 \text{ years} \approx 0.79 \text{ trading days} \]

This suggests rebalancing slightly more frequently than once per day.

The minimum total cost is:

\[ \text{Total}(N^*) = \frac{a}{\sqrt{a/b}} + b\sqrt{a/b} = \sqrt{ab} + \sqrt{ab} = 2\sqrt{ab} \]

\[ = 2\sqrt{4.0 \times 0.05} = 2\sqrt{0.20} = 2 \times 0.4472 \approx 0.894 \]

The minimum total cost is approximately \(\$0.89\).

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Exercise 3. Leland's adjusted volatility is \(\hat{\sigma}^2 = \sigma^2(1 + \sqrt{2/\pi} \cdot \kappa/(\sigma\sqrt{\delta t}))\). For \(\sigma = 0.20\), \(\kappa = 0.002\) (20 bps half-spread), and daily rebalancing (\(\delta t = 1/252\)), compute \(\hat{\sigma}\). By what percentage does the Leland volatility exceed the true volatility? How does this percentage change if the trader switches to weekly rebalancing?

Solution to Exercise 3

With \(\sigma = 0.20\), \(\kappa = 0.002\), \(\delta t = 1/252\):

\[ \hat{\sigma}^2 = \sigma^2\left(1 + \sqrt{\frac{2}{\pi}} \cdot \frac{\kappa}{\sigma\sqrt{\delta t}}\right) \]

Compute the correction factor:

\[ \frac{\kappa}{\sigma\sqrt{\delta t}} = \frac{0.002}{0.20 \times \sqrt{1/252}} = \frac{0.002}{0.20 \times 0.06299} = \frac{0.002}{0.01260} \approx 0.15873 \]

\[ \sqrt{\frac{2}{\pi}} \approx 0.7979 \]

\[ \hat{\sigma}^2 = 0.04\left(1 + 0.7979 \times 0.15873\right) = 0.04(1 + 0.12665) = 0.04 \times 1.12665 = 0.04507 \]

\[ \hat{\sigma} = \sqrt{0.04507} \approx 0.2123 \]

The Leland volatility exceeds the true volatility by:

\[ \frac{\hat{\sigma} - \sigma}{\sigma} = \frac{0.2123 - 0.20}{0.20} = \frac{0.0123}{0.20} \approx 6.15\% \]

For weekly rebalancing (\(\delta t = 1/52\)):

\[ \frac{\kappa}{\sigma\sqrt{\delta t}} = \frac{0.002}{0.20 \times \sqrt{1/52}} = \frac{0.002}{0.20 \times 0.13868} = \frac{0.002}{0.02774} \approx 0.07211 \]

\[ \hat{\sigma}^2 = 0.04(1 + 0.7979 \times 0.07211) = 0.04(1 + 0.05753) = 0.04 \times 1.05753 = 0.04230 \]

\[ \hat{\sigma} \approx \sqrt{0.04230} \approx 0.2057 \]

The percentage excess is:

\[ \frac{0.2057 - 0.20}{0.20} \approx 2.85\% \]

The Leland adjustment is smaller for weekly rebalancing (\(2.85\%\) vs \(6.15\%\)) because less frequent rebalancing incurs fewer transaction costs per unit time, requiring a smaller volatility correction.

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Exercise 4. The expected total transaction cost scales as \(\mathbb{E}[\text{TC}_{\text{total}}] \approx C_0 \sqrt{N}\). For an ATM call with \(\Gamma = 0.032\), \(\sigma = 0.20\), \(S = 100\), \(T = 0.25\), and \(\kappa = 0.001\), compute \(C_0\) and the expected transaction costs for daily and hourly rebalancing. At what rebalancing frequency do expected transaction costs equal \(1\%\) of the option price (approximately \(\$4.50\))?

Solution to Exercise 4

From the formula \(\mathbb{E}[\text{TC}_{\text{total}}] \approx C_0\sqrt{N}\) where:

\[ C_0 = \kappa\,\overline{|\Gamma|\sigma S^2}\,T\,\sqrt{\frac{2}{\pi}} \]

Wait -- more precisely, each step costs \(\mathbb{E}[\text{TC}_k] \approx \kappa\,|\Gamma_k|\,\sigma S_k^2\sqrt{2\delta t/\pi}\), and summing \(N\) steps:

\[ \mathbb{E}[\text{TC}_{\text{total}}] = N \times \kappa\,|\Gamma|\,\sigma S^2\sqrt{\frac{2\delta t}{\pi}} = \kappa\,|\Gamma|\,\sigma S^2\,\sqrt{\frac{2N}{\pi}} \cdot \sqrt{T} \]

Let us define \(C_0\) so that \(\mathbb{E}[\text{TC}_{\text{total}}] = C_0\sqrt{N}\):

\[ C_0 = \kappa\,|\Gamma|\,\sigma S^2\,\sqrt{\frac{2T}{\pi}} \]

With \(\kappa = 0.001\), \(\Gamma = 0.032\), \(\sigma = 0.20\), \(S = 100\), \(T = 0.25\):

\[ C_0 = 0.001 \times 0.032 \times 0.20 \times 10000 \times \sqrt{\frac{0.5}{\pi}} \]

\[ = 0.001 \times 0.032 \times 0.20 \times 10000 \times \sqrt{0.15915} \]

\[ = 0.064 \times 0.3989 \approx 0.02553 \]

Daily rebalancing (\(N = 63\) for \(T = 0.25\)):

\[ \mathbb{E}[\text{TC}] = 0.02553 \times \sqrt{63} = 0.02553 \times 7.937 \approx 0.2026 \]

Hourly rebalancing (\(N = 63 \times 8 = 504\)):

\[ \mathbb{E}[\text{TC}] = 0.02553 \times \sqrt{504} = 0.02553 \times 22.45 \approx 0.573 \]

Frequency where TC equals 1% of option price. With option price \(\approx \$4.50\), we need \(\text{TC} = 0.045\):

\[ 0.02553\sqrt{N} = 0.045 \implies \sqrt{N} = 1.763 \implies N \approx 3.1 \]

Transaction costs reach \(1\%\) of the option price at approximately \(N = 3\) rebalancing events over the life of the option, which corresponds to monthly rebalancing (\(\delta t \approx 0.25/3 \approx 0.083\) years). For any higher frequency, TC exceeds \(1\%\) of the option price.

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Exercise 5. Using the CLT for hedging error, construct a 95% confidence interval for the cumulative hedging error of a daily-rebalanced short ATM call with \(S = K = 100\), \(\sigma = 0.20\), \(T = 0.5\), \(\Gamma = 0.025\). Express the interval in dollars and as a percentage of the option price (approximately \(\$6.30\)). How does the interval change if gamma-weighted adaptive rebalancing reduces the effective number of high-gamma steps by half?

Solution to Exercise 5

With \(S = K = 100\), \(\sigma = 0.20\), \(T = 0.5\), \(\Gamma = 0.025\), daily rebalancing (\(\delta t = 1/252\)):

\[ \operatorname{Var}(\text{HE}) = \frac{1}{2}\Gamma^2 S^4 \sigma^4 T \cdot \delta t \]

\[ = \frac{1}{2}(0.025)^2(10^8)(0.20)^4(0.5)(1/252) \]

\[ = \frac{1}{2}(6.25 \times 10^{-4})(10^8)(1.6 \times 10^{-3})(0.5)(3.968 \times 10^{-3}) \]

\[ = \frac{1}{2}(6.25 \times 10^{-4})(10^8)(3.175 \times 10^{-6}) \]

\[ = \frac{1}{2}(6.25 \times 10^{-4})(317.5) = \frac{1}{2}(0.1984) = 0.09922 \]

\[ \operatorname{Std}(\text{HE}) = \sqrt{0.09922} \approx 0.3150 \]

The 95% confidence interval is:

\[ [-1.96 \times 0.315,\; +1.96 \times 0.315] = [-0.617,\; +0.617] \]

In dollars: the interval is approximately \([-\$0.62, +\$0.62]\).

As a percentage of option price (\(\approx \$6.30\)):

\[ \frac{0.617}{6.30} \approx 9.8\% \]

The 95% confidence interval is approximately \(\pm 9.8\%\) of the option price.

With adaptive rebalancing reducing effective high-gamma steps by half: the dominant contribution to \(\operatorname{Var}(\text{HE})\) comes from high-gamma steps. If adaptive scheduling halves the effective number of these steps (by rebalancing more frequently during high-gamma periods and less frequently otherwise), the variance decreases by roughly a factor of 2:

\[ \operatorname{Std}(\text{HE})_{\text{adaptive}} \approx \frac{0.315}{\sqrt{2}} \approx 0.223 \]

The 95% CI shrinks to approximately \([-\$0.44, +\$0.44]\), or about \(\pm 6.9\%\) of the option price.

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Exercise 6. A trader considers a non-uniform rebalancing schedule where the interval shrinks near expiry. The adaptive rule triggers rebalancing when \(\frac{1}{2}|\Gamma_k| \sigma^2 S_k^2 (t - t_k) > \varepsilon_{\text{target}}\). For \(\varepsilon_{\text{target}} = 0.05\), \(\sigma = 0.20\), \(S = 100\), and \(\Gamma\) ranging from \(0.03\) (at \(\tau = 3\) months) to \(0.30\) (at \(\tau = 1\) day), compute the maximum allowable interval \(\delta t_{\max}\) at each gamma level. How does this compare to a uniform daily schedule?

Solution to Exercise 6

The adaptive rule triggers rebalancing when:

\[ \frac{1}{2}|\Gamma_k|\,\sigma^2 S_k^2\,(t - t_k) > \varepsilon_{\text{target}} \]

Solving for the maximum allowable interval:

\[ \delta t_{\max} = \frac{2\varepsilon_{\text{target}}}{|\Gamma_k|\,\sigma^2 S_k^2} \]

With \(\varepsilon_{\text{target}} = 0.05\), \(\sigma = 0.20\), \(S = 100\):

\[ \delta t_{\max} = \frac{2 \times 0.05}{|\Gamma_k| \times 0.04 \times 10000} = \frac{0.10}{400\,|\Gamma_k|} \]

At \(\tau = 3\) months (\(\Gamma = 0.03\)):

\[ \delta t_{\max} = \frac{0.10}{400 \times 0.03} = \frac{0.10}{12} \approx 0.00833 \text{ years} \approx 2.1 \text{ trading days} \]

At \(\tau = 1\) day (\(\Gamma = 0.30\)):

\[ \delta t_{\max} = \frac{0.10}{400 \times 0.30} = \frac{0.10}{120} \approx 0.000833 \text{ years} \approx 0.21 \text{ trading days} \approx 1.7 \text{ hours} \]

Comparison with uniform daily schedule (\(\delta t = 1/252 \approx 0.00397\) years):

Time to expiry	\(\Gamma\)	\(\delta t_{\max}\)	Uniform daily	Comparison
3 months	0.03	2.1 days	1 day	Adaptive allows longer intervals
1 day	0.30	1.7 hours	1 day	Adaptive requires much shorter intervals

When gamma is low (far from expiry), the adaptive rule permits less frequent rebalancing than daily, saving transaction costs. Near expiry, when gamma spikes, the adaptive rule demands sub-daily rebalancing to control the hedging error. A uniform daily schedule over-hedges early (wasting on transaction costs) and under-hedges near expiry (accepting too much error). The adaptive approach allocates rebalancing effort where it matters most.